288KJ is conducted across two opposite faces of a 3m cube of temperature gradient 90ºCm−1 − 1 in 7200s. Calculate the thermal conductivity.

The thermal conductivity of a material is a measure of its ability to conduct heat. It is defined by the formula:

Q = k × A × ΔT/Δx × t

Where:

• Q is the heat energy conducted (in Joules)
• k is the thermal conductivity (in W/mK)
• A is the area through which heat is conducted (in square meters, m²)
• ΔT is the temperature difference (in Kelvin or Celsius, since a difference in degrees Celsius is equivalent to a difference in Kelvin)
• Δx is the thickness of the material (or the distance over which the temperature difference occurs, in meters)
• t is the time taken (in seconds)

We are given:

• Q = 288,000 J (since energy is in KJ and 1KJ = 1,000J)
• ΔT/Δx = 90 ºC/m (temperature gradient)
• t = 7,200 s

The cube has each side measuring 3 meters, so the area A of one face (since heat is conducted across two opposite faces, effectively using one face area for calculation) is:

A = 3m × 3m = 9 m²

Now, we need to solve for k (thermal conductivity):

Q = k × A × ΔT/Δx × t

288,000 J = k × 9 m² × 90 ºC/m × 7,200 s

k = 288,000 / (9 × 90 × 7,200)

Calculate the denominator:

9 × 90 × 7,200 = 5,832,000

Therefore:

k = 288,000 / 5,832,000 ≈ 0.0493 W/mK

This converts approximately to 4.93 × 10^-2 W/mK.

Therefore, the correct answer is 4.9 × 10^-2 W/mK.